Question

An object of mass 0.2 kg executes simple harmonic motion along X-axis with frequency of 25/ $\pi$ Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation in meter is equal to

• A. 0.05 m
• B. 0.06 m
• C. 0.01 m
• D. 0.002 m

Answer 1

The correct option is B

0.06 m

$E=\frac{1}{2}m{\omega }^2{A}^2$

$\Rightarrow E=\frac{1}{2}m(2\pi f{)}^2{A}^2$

$\Rightarrow A=\frac{1}{2\pi f}\sqrt{\frac{2E}{m}}$

Putting E = K + U we obtain, $A=\frac{1}{2\pi \left(\frac{25}{\pi }\right)}\sqrt{\frac{2\times (0.5+0.4)}{0.2}}$

$\Rightarrow A=0.06m.$